How do you differentiate # f(x)= sin(x-2)^3 # using the chain rule.?

1 Answer
Feb 2, 2016

Answer:

#f'(x)=3(x-2)^2cos[(x-2)^3]#

Explanation:

The chain rule states that if #y=f(u) and u=f(x)# then
#dy/dx=(dy)/(du)*(du)/dx#

Furthermore, the power rule states that
#d/dx[u(x)]^n=n*[u(x)]^(n-1)*(du)/dx#

And then finally we also have that #d/dxsin[u(x)]=cosu*(du)/dx#.

So application of these rules yields the following result :

#d/dx[sin(x-2)^3]=cos[(x-2)^3]*3(x-2)^2*1#

#=3(x-2)^2cos[(x-2)^3]#